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Grade 12th PassMechanics

two particles of equal mass m go around a circle of radius R under the influence of their mutual gravitational attraction.the speed of each particle with respect to their center of mass will be??

Profile image of bijitesh choudhury
13 Years agoGrade 12th Pass
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7 Answers

Profile image of rohit yadav
13 Years ago

with respect to the center of mass the speed of each particle is (Gm/4R)1/2 as their is no external force acting on the system so velocity of the center of mass is zero velocity of each particle is (Gm/4R)1/2.

Profile image of Chetan Mandayam Nayakar
13 Years ago

mv2/R=Gm2/(2R)2

Profile image of Yogita Bang
13 Years ago

The gravitational force of attraction between the two masses is

F = Gm2/(2R)2

This force provides the necessary centripetal force

mv2/R = Gm2/4R2

v2 = Gm/4R

v = √(Gm/4R)

Profile image of gauhar singh
13 Years ago

The Speed will be equal to their velocities in inertial frame itself as there centre of mass will be stationary.

Profile image of Ojas Suvarnakar
9 Years ago
Here center of mass will be stationary because both particles are having same mass,Therefore ,Speed of the particle will be as equal as critical velocity of the particle...Therefore answer is √GM/R
Profile image of Aniket lohar
8 Years ago
Two particles of equal mass(m) go around a circle of radius R under the action of their mutual gravitational attraction.Here center of mass is stationary because both particles are having same mass.1/2√(GM/R)
Profile image of Yug Dedhia
7 Years ago
Hello this is master Yug Dedhia
Here,using theory of gravitation
Therefore,mv^2/ r = Gm^2/(2R)^2
Thus the answer is v=√(Gm/4R)
Thanks